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This might be a very simple question, but: a class of 25 students is divided into 5 teams of 5 each. What is the probability of student X and Y being in the same team?

is it just 4/25?

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It is $4/24$.${}{}$ – André Nicolas Feb 25 '14 at 0:33
Supose X is in a class....there are 24 remaining students and 4 remaining slots... – Faffi-doo Feb 25 '14 at 0:35
ahhh okok yes it's 4/24. alright thanks! – John Feb 25 '14 at 0:36
up vote 3 down vote accepted

Imagine that our heroes, A and B, are assigned to teams, in that order, with the rest being assigned later.

Whatever team A is assigned to, there are $4$ empty spots on that team. The probability B is given one of these spots is $\frac{4}{24}$.

Or else we can do more elaborate counting. Imagine the teams are labelled (it makes no difference to the probability). There are $\binom{25}{5}\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5}$ ways to assign the people to labelled teams, all equally likely.

Now we count the number of ways A and B can end up on the same team. Which team? It can be chosen in $5$ ways. The other $3$ people on that team can be chosen in $\binom{23}{3}$ ways. And the rest of the assignments can be done in $\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5}$ ways.

Divide. We get that the probability is $\frac{5\binom{23}{3}}{\binom{25}{5}}$. This simplifies to $\frac{1}{6}$.

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Hint. Once you know which team X is on, how many places altogether are available for Y? And how many of those available places are in the same team as X?

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